\section{The degree and dimension induction}\label{sec:induction}

\subsection{The range covered by one hyperplane}

\begin{proposition}\label{prop:small-dimension-range}
Let $D\geq4$, and suppose that the ideal generation statement holds in
degrees $D-1$ and $D-2$ for every number of variables.  Then it holds
in degree $D$ for
\[
                 2\leq n\leq D(D-1)+1.
\]
\end{proposition}
\begin{proof}
We use induction on $n$ at fixed degree $D$. The case $n=2$ follows
from the principal ideal description on $\mathbf P^1$, and $n=3$
from Lemma~\ref{lem:plane-base}. Thus the same-degree trace
statement is available whenever an integral or fractional transfer is
applied below.

When $E(n-1,D)$ is integral, the
integral transfer asks for injectivity at residual dimension
\[
       \left\lceil\frac{N_{D-1}(n)}{D+1}\right\rceil.
\]
For $n\geq4$,
\[
 \frac{N_{D-1}(n)}{D(D+1)}
 \geq\frac{N_{D-1}(4)}{D(D+1)}
 =\frac{D+2}{6}\geq1.
\]
Consequently the required integer is at most
$N_{D-1}(n)/D=E(n,D-1)$.  The preceding-degree hypothesis gives
injectivity.  This residual lies in the ordinary scalar postulation
cell, since $E(n,D-1)\leq N_{D-1}(n-1)$.

For $D=4$, the only fractional trace edges in the range $4\le n\le13$
are $n=7,12$: the integer $\binom{n+2}{4}$ is divisible by $5$
unless $n\equiv2\pmod5$. The parameters of
Proposition~\ref{prop:fractional-trace} are
\[
\begin{array}{c|rrrrrrrr}
 n&\tau&D_1&q_1&b_1&b_0&a=a'&\rho_I=\rho_S&k=\tau+q_1\\ \hline
 7 &25&84&16&4&4&0&3&41\\
 12&200&143&14&3&8&0&2&214
\end{array}
\]
Both rows have $D_1\ge0$ and $k\ge(n-1)\rho_S$.
The three auxiliary counts are $41,42,43$ in the cubic point
interval $[N_2(6),N_3(6)-1]=[21,55]$, and $214,215,216$ in
$[N_2(11),N_3(11)-1]=[66,285]$, respectively.
The residual quadratic ideal dimensions are $3$ and $2$, below
$E(7,2)=28/3$ and $E(12,2)=26$. Their point counts are $25$ and
$76$, in the quadratic initial-degree intervals; in particular
$1\le\rho_I=\rho_S\le N_2(n-1)$.
Thus the two preceding-degree hypotheses and the smaller
same-degree trace supply every hypothesis of the fractional
transfer. Together with the integral steps, this proves the
proposition for $D=4$.

Assume henceforth that $D\ge5$. We first treat fractional trace edges
with $n\ge5$. Use the notation
$b_0,\tau,D_1,q_1,b_1,a,a',\rho_I,\rho_S$ of the fractional transfer,
and put
\[
 \kappa=\frac{(n+D-1)(n+D-2)}{D(D+1)},
 \qquad c=D(D-1)+1-n.
\]
The identity
\begin{equation}\label{eq:small-range-capacity}
 D_1=N_{D-1}(n-1)(n-1-\kappa)
     =\frac{N_{D-1}(n-1)(n-2)c}{D(D+1)}
\end{equation}
shows that $D_1\geq0$ throughout the asserted range.

The scalar-row budget follows from
\[
 (n-1)\rho_S=q_1+n-1-b_1+a'\leq q_1+2n-3
 \quad\hbox{and}\quad \tau\geq2n-3.
\]
For the latter inequality, monotonicity in $D$ gives
\[
 E(n-1,D)\geq E(n-1,4)=\frac15\binom{n+2}{4}.
\]
At $n=5$ this equals $2n-3=7$, and its increment as $n$ increases
is $\binom{n+2}{3}/5>2$.

The residual dimension is strictly below its critical value.
Indeed, the definitions and $0\leq a'\leq n-2$ give
\begin{align*}
 \rho_S
 &<E(n,D-2)\frac{n-1-\kappa}{n-2}+2\\
 &=E(n,D-2)-E(n,D-2)\frac{\kappa-1}{n-2}+2.
\end{align*}
Here
\[
 \frac{\kappa-1}{n-2}
   =\frac{2D+n-1}{D(D+1)}\geq\frac2{D+1},
 \qquad
 E(n,D-2)\geq E(5,D-2)
   =\frac{D(D+1)(D+2)}{24}\geq D+1.
\]
It follows that $\rho_S<E(n,D-2)$.  The degree-$(D-2)$ hypothesis
therefore supplies residual injectivity, once positivity is checked.

At the upper endpoint $n=D(D-1)+1$, the trace edge is integral.
To see this, write
\[
 D\binom{D^2-1}{D}
   =(D^2-1)\binom{D^2-2}{D-1}.
\]
The right side is divisible by $D+1$, and $\gcd(D,D+1)=1$.
Thus a fractional edge has $c\geq1$.  We claim that
\begin{equation}\label{eq:small-range-positive}
       N_{D-1}(n-1)c\geq D(D+1)(n-2),
       \qquad q_1\geq n-2.
\end{equation}
The second assertion follows from the first and
\eqref{eq:small-range-capacity}.  If $5\leq n\leq D+2$, the ratio
$N_{D-1}(n-1)/(n-2)$ increases with $n$, so
\[
 \frac{N_{D-1}(n-1)c}{n-2}
 \geq\frac{D(D+1)(D+2)}{18}(D^2-2D-1)
 \geq D(D+1).
\]
If $n\geq D+3$, use $c\geq1$ and
$N_{D-1}(n-1)\geq\binom{n+2}{4}$.  The ratio
$\binom{n+2}{4}/(n-2)$ increases with $n$.  At $n=D+3$ the required
inequality is
$\binom{D+5}{4}\geq D(D+1)^2$; after multiplication by $24$, its
difference is
\[
 (D-5)^4+10(D-5)^3+23(D-5)^2+110(D-5)+720>0.
\]
This proves \eqref{eq:small-range-positive}.  Since
$a\leq n-2$ and $b_1\leq n-3$,
\[
 (n-1)\rho_I=q_1+n-1-a-b_1\geq2.
\]
Both residual dimensions are therefore positive integers, and
$\rho_I\leq\rho_S<E(n,D-2)$.  Also
$E(n,D-2)\leq N_{D-2}(n-1)$, which gives the required scalar cell
and the vanishing in the preceding degree.

It remains to check the three auxiliary trace counts.  Set
$k=\tau+q_1$ and $r'=N_{D-1}(n-1)-k$.  The exact identity
\[
 r'-E(n-1,D-1)=1+\frac{b_1-b_0}{n-2}
\]
places $r'$ strictly between $E(n-1,D-1)$ and
$E(n-1,D-1)+2$.  Moreover,
\[
 E(n-1,D-1)\geq E(4,4)=7,
 \qquad
 E(n-1,D-1)\leq\tfrac12N_{D-1}(n-2).
\]
The last inequality is equivalent to $(D-2)(n-4)\geq2$.
Since $N_{D-1}(n-2)\geq15$, all three integers
$r',r'-1,r'-2$ lie between $1$ and $N_{D-1}(n-2)$.
Equivalently, $k,k+1,k+2$ lie in the required point cell.
Every hypothesis of the fractional transfer is now satisfied.

For completeness, consider $n=4$.  A fractional trace edge occurs
exactly when $D$ is odd.  For odd $D\geq7$, direct substitution gives
\[
 \begin{gathered}
 b_0=b_1=1,\qquad
 \tau=\frac{D+1}{2},\qquad
 D_1=D^2-D-3,\qquad q_1=\frac{D^2-D-4}{2},\\
 \rho_I=\frac{D(D-1)}6-\frac a3,
 \qquad
 \rho_S=\frac{D(D-1)}6+\frac{a'}3,
 \end{gathered}
\]
where $(a,a')$ is either $(0,0)$ or $(1,2)$.
Thus $\rho_I\geq1$, and
\[
 \rho_S<E(4,D-2)=\frac{D(D+1)}6.
\]
The scalar-row budget reduces to
$\tau\geq2+a'$, which holds for $D\geq7$.
The three auxiliary ideal dimensions are
$(D+3)/2,(D+1)/2,(D-1)/2$; they all lie in $[1,D]$.
The fractional transfer applies.  The only remaining case is
$(n,D)=(4,5)$.  Apply the ordinary splitting lemmas with residual
dimension $r_1=6$ and trace dimensions $r_2=3$ and $r_2=4$,
respectively.  The residual is injective because
$6\leq E(4,4)=7$, while
\[
 3\cdot6+3=21=N_5(3),\qquad 3\cdot6+4\geq N_5(3).
\]
The resulting dimensions $9$ and $10$ are precisely the two critical
dimensions in degree five.  This completes the proof for $D\geq5$.

\end{proof}

\subsection{Completion of the proof}

\begin{proof}[Proof of Theorem~\ref{thm:igc}]
The quadratic and cubic cases are Theorems~\ref{thm:quadratic-column}
and \ref{thm:cubic}. For $D\ge4$, assume that the statement holds in
degrees $D-1$ and $D-2$ for every number of variables.
Proposition~\ref{prop:small-dimension-range} proves the degree-$D$
statement for $n\le D(D-1)+1$.

Proceed by induction on $n$ beyond this range. Put $m=n-2$; then
$m\ge D(D-1)$, and the degree-$D$ statement in $m$ variables is
already known. If $E(m,D)$ is integral, apply
Proposition~\ref{prop:integral-two-normal}; otherwise apply
Proposition~\ref{prop:fractional-two-normal}.
Their two lower-degree hypotheses are precisely the induction assumptions.
In both cases the two critical counts in $n$ variables have maximal
rank. Lemma~\ref{lem:critical-counts} gives every point count in the
initial-degree interval. This completes the induction on $n$ and
then on $D$.
\end{proof}